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Consider simple, undirected Erdős–Rényi graphs $G(n,p)$, where $n$ is the number of vertices and $p$ is the probability for each pair of vertices to form an edge. Many properties of these graphs are known - in particular, $G(n,p)$ is almost surely connected when $p \gt (1 + \epsilon)\frac{\log(n)}{n}$, and the largest clique in $G(n, \frac{1}{2})$ is almost surely about $2\log_2(n)$.

What is known about the vertex connectivity number $\kappa(G)$, $G\in G(n,p)$, the minimum number of vertices that one must remove in order to disconnect the graph?

It is known that for fixed $k$ and fixed $p\in (0,1)$, almost every graph in $G(n,p)$ is $k$-connected, but what is the expected connectivity as a function of $p$ and $n$?

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  • $\begingroup$ Yes, you are correct. $\endgroup$ Commented Jun 16, 2010 at 0:42

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The expected connectivity cannot be higher than the expected minimal degree, which jumps to roughly $pn$ after getting into the range $p\gg\frac{\log n}{n}$. On the other hand, sloppily counting potential clusters of size $m < n/2$ that have boundaries of less than $k$ vertices gives a probability of $\binom{n}{m}\binom{n-m}{k}(1-p)^{m(n-m-k)}$, which is for $k \ll n$ decreasing in $m$ up to $m\approx \frac{n-k}{2}$ and increasing after that value, so we can get an estimate by considering only $m=1$ (checking for vertices with at most $k$ neighbours) and $m=\frac{n}{2}$: $$ \binom{n}{n/2}\binom{n/2}{k}(1-p)^{n(n-2k)/4} < \exp(n \log 2+k \log n - pn(n-2k)/4) < $$ $$ < \exp(n \log 2 - pn(\frac{n}{4}-\frac{k}{2}-\log n)) < \exp(-\frac{n \log n}{4} + n \log 2 +2(\log n)^2), $$ this latter number tending to $0$ fast enough to ignore it. So, the expected connectivity is the expected minimal degree and is roughly $pn$ once $p$ exceeds $\log n/n$. Do you need the behaviour of expected connectivity specifically in this region?

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  • $\begingroup$ Thanks Thorny - this was very helpful. So the $E(\kappa(H))$ tends to around $\frac{n}{2}$ for $H \in G(n,p)$, which is what I was most interested in. $\endgroup$ Commented Jun 16, 2010 at 15:08

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